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Westkost [7]
3 years ago
6

you draw a card from a deck if you get a red card you win nothing if you get a spade you win 14 for any club you win 28 plus an

extra 10 for the ace of clubs find the stanard deviation of the amount you might win drawin a card
Mathematics
1 answer:
Readme [11.4K]3 years ago
6 0

Answer: Standard deviation is 11.6

Step-by-step explanation:

Sample mean 52/4=13

X. X---x(bar). X---x(bar)^2

0. 0--13=-13. -13^2=169

14. 14--13=1. 1^2=1

28. 28--13=15. 15^2=225

10. 10--13=--3. --3^2=9

__________________________

52. 404

Standard deviation=√404/3=11.6

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<h3>When do we perform one sample z-test?</h3>

One sample z-test is performed if the sample size is large enough (n  > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

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  • Level of significance = \alpha = 2.5% = 0.025
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Forming hypotheses:

  • Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: H_0: \mu_0 \leq \mu = 150
  • Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically H_1: \mu_0 > \mu = 150

where \mu_0 is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

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Learn more about one-sample z-test here:

brainly.com/question/21477856

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<h3><u>Solution:</u></h3>

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<em><u>Another way:</u></em>

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